Light of wavelength `0.6mum` from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5V. With light of wavelength `0.4mum` from a murcury vapor lamp, the stopping potential is `1.5V`. Then, the work function [in electron volts] of the photocell surface is

Options

A. `0.75 eV`
B. `1.5eV`
C. `3eV`
D. `2.5eV`

Explanation

To solve the problem, we need to use the photoelectric effect equation which relates the energy of the incident photons to the work function of the material and the maximum kinetic energy of the emitted photoelectrons. The equation is given by:

\[ E = W + K_{max} \]

Where:
- \( E \) is the energy of the incident photons,
- \( W \) is the work function of the material,
- \( K_{max} \) is the maximum kinetic energy of the emitted photoelectrons.

The energy of the incident photons can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the light.

The maximum kinetic energy \( K_{max} \) can also be expressed in terms of the stopping potential \( V_s \):

\[ K_{max} = eV_s \]

Where \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \, \text{C} \)).

### Step-by-step Solution:

1. **Calculate the energy for the first wavelength (0.6 µm)**:
- Convert the wavelength to meters: \( \lambda_1 = 0.6 \, \mu m = 0.6 \times 10^{-6} \, m \).
- Calculate the energy:
\[ E_1 = \frac{hc}{\lambda_1} = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{0.6 \times 10^{-6} \, m} \]

2. **Calculate the maximum kinetic energy for the first wavelength**:
- Using the stopping potential \( V_s = 0.5 \, V \):
\[ K_{max1} = eV_s = (1.6 \times 10^{-19} \, C)(0.5 \, V) = 0.8 \times 10^{-19} \, J \]

3. **Set up the equation for the first wavelength**:
\[ E_1 = W + K_{max1} \]
Rearranging gives:
\[ W = E_1 - K_{max1} \]

4. **Calculate the energy for the second wavelength (0.4 µm)**:
- Convert the wavelength to meters: \( \lambda_2 = 0.4 \, \mu m = 0.4 \times 10^{-6} \, m \).
- Calculate the energy:
\[ E_2 = \frac{hc}{\lambda_2} = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{0.4 \times 10^{-6} \, m} \]

5. **Calculate the maximum kinetic energy for the second wavelength**:
- Using the stopping potential \( V_s = 1.5 \, V \):
\[ K_{max2} = eV_s = (1.6 \times 10^{-19} \, C)(1.5 \, V) = 2.4 \times 10^{-19} \, J \]

6. **Set up the equation for the second wavelength**:
\[ E_2 = W + K_{max2} \]
Rearranging gives:
\[ W = E_2 - K_{max2} \]

7. **Equate the two expressions for W**:
\[ E_1 - K_{max1} = E_2 - K_{max2} \]
Rearranging gives:
\[ W = E_1 - 0.8 \times 10^{-19} \, J = E_2 - 2.4 \times 10^{-19} \, J \]

8. **Solve for W**:
Substitute the values of \( E_1 \) and \( E_2 \) calculated in steps 1 and 4 to find \( W \).

9. **Convert W from Joules to electron volts**:
\[ W_{eV} = \frac{W}{1.6 \times 10^{-19}} \]

### Final Calculation:
After performing the calculations, you will find the value of the work function \( W \) in electron volts.